Oracle按时间段分组统计

想要按时间段分组查询，首先要了解level,connect by,oracle时间的加减.

关于level这里不多说，我只写出一个查询语句:

----level 是一个伪例
select level from dual connect by level <=10
---结果：
1
2
3
4
5
6
7
8
9
10

关于connect by可以看

http://www.cnblogs.com/johnnyking39/articles/1155497.html

oracle时间的加减看看试一下以下sql语句就会知道:

select sysdate -1 from dual
----结果减一天,也就24小时
select sysdate-(1/2) from dual
-----结果减去半天，也就12小时
select sysdate-(1/24) from dual
-----结果减去1 小时
select sysdate-((1/24)/12) from dual
----结果减去5分钟
select sydate-(level-1) from dual connect by level<=10
---结果是10间隔1天的时间

下面是本次例子：

select dt, count(satisfy_degree) as num from T_DEMO  i ,
(select sysdate - (level-1) * 2 dt
from dual connect by level <= 10) d
where i.satisfy_degree='satisfy_1' and
i.insert_time<dt and i.insert_time> d.dt-2
group by d.dt

例子中的sysdate - (level-1) * 2得到的是一个间隔是2天的时间

group by d.dt 也就是两天的时间间隔分组查询

自己实现例子：

create table A_HY_LOCATE1
(
MOBILE_NO          VARCHAR2(32),
LOCATE_TYPE        NUMBER(4),
AREA_NO            VARCHAR2(32),
CREATED_TIME       DATE,
AREA_NAME          VARCHAR2(512),
)；

select (sysdate-13)-(level-1)/4 from dual connect by level<=34 --从第一条时间记录开始（sysdate-13）为表中的最早的日期，“34”出现的分组数（一天按每六个小时分组 就应该为4）

一下是按照每6个小时分组

select mobile_no,area_name,max(created_time ),dt, count(*) as num from a_hy_locate1  i ,
(select (sysdate-13)-(level-1)/4  dt
from dual connect by level <= 34) d
where i.locate_type = 1 and
i.created_time<dt and i.created_time> d.dt-1/4
group by mobile_no,area_name,d.dt

另外一个方法：

--按六小时分组
select trunc(to_number(to_char(created_time, 'hh24')) / 6),count(*)
from t_test
where created_time > trunc(sysdate - 40)
group by trunc(to_number(to_char(created_time, 'hh24')) / 6)

--按12小时分组
select trunc(to_number(to_char(created_time, 'hh24')) / 6),count(*)
from t_test
where created_time > trunc(sysdate - 40)
group by trunc(to_number(to_char(created_time, 'hh24')) / 6)

出处：http://blog.csdn.net/wanglipo/article/details/6556665